LECTURE 1 Introduction to Mechanics of Materials. Geometrical Properties of Cross Sections of a Rod (Part 1)

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1 V. DEMENKO MECHNCS OF MTERLS 05 LECTURE ntroduton to Mehans of Materals. Geometral Propertes of Cross Setons of a Rod (Part ) Prolems and Methods n Mehans of Materals (syn. Strength of Materals) Mehans of materals s the sene of strength stffness and stalty of elements of engneerng strutures. Strength s understood as the alty of struture or ts elements to wthstand a spefed external loadng wthout frature. Stffness (or rgdty) s understood as the apalty of a ody or strutural element to resst deformaton.e. to prevent exeedng elongatons defletons and so on. Stalty s meant as the apalty of a struture to resst the fores whh tend to move t from the ntal state of equlrum.e. to prevent uklng. Mehans of materals s one of the ranhes of mehans of deformale solds. Mehans of deformale solds nludes also other ranhes suh as the mathematal theory of elastty theory of plates and shells. The mathematal theory of elastty studes the ehavor of deformale solds usng a omplex mathematal apparatus. Strength of materals uses a smple mathematal apparatus and smplfyng hypotheses. t reates smple approxmate alulaton of typal strutural elements for strength rgdty and stalty. The struture sn't ale to work at the level of frature.e. should not fal under appled external loads. t must have prelmnary grounded fator of safety. The lak of fator leads to frature ut nsuffent fator makes struture mperfet. The orret hoe of fator s a responsle prolem n mehanal engneerng. The geometral sheme n strength of materals s the sheme of a rod. rod generally mples a ody one of whose dmensons (length) s onsderaly greater than the materals. other two. Bars eams shafts shells are also onsdered n mehans of

2 V. DEMENKO MECHNCS OF MTERLS 05 Geometral Propertes of Cross Setons of a Rod n solvng the prolems n strength of materals t s neessary to operate wth some geometral propertes of ross setons of a rod whh nfluene on alty of engneerng struture to wthstand appled load. Smplest example of a rod proessng y weldng s shown n Fg... Cross-Seton rea Fg. Fg. Take a ross seton of a rod. Relate t to a system of oordnates y z. solate an element from the area wth oordnates y z. Consder the followng ntegral: = lm = d 0 () where the ndex eneath the ntegral sgn ndates that the ntegraton s arred out over the whole ross-setonal area. The ntegral () s alled as ross-seton area. Cross-setonal areas of smple fgures Сrle πd = πr = area. Сrular setor α =angle n radans ( α π ) = αr area. Fg. Fg.

3 V. DEMENKO MECHNCS OF MTERLS 05 Сrular segment Orgn of axes at enter of rle α π α =angle n radans ( ) r ( α snαosα) = area. Crle wth ore removed α =angle n radans ( α π ) aros a α = = r a ; r a = r α r area. Fg. 5 Fg. 6 Сrular segment Orgn of axes at enter of rle α π α =angle n radans ( ) r ( α snαosα) = area. Equlateral trangle a sde = a area. Fg. 7 Fg.

4 V. DEMENKO MECHNCS OF MTERLS 05 Ellpse Orgn of axes at entrod = π a a magor axs mnor axs; Crumferene π.5 ( a ) a + ( a/ a) ( ).7 / a+ a 0 a/. Hollow rular ross seton ( ) = π r r r nner radus r outer radus t = r r thkness. Fg. 9 Fg. 0 Hollow square ross seton (douly symmetr) = area C entrod. soseles trapezod ( + ) h = area C entrod h heght. Fg. Fg.

5 V. DEMENKO MECHNCS OF MTERLS 05 5 soseles rght trangle = area C entrod. soseles trangle = h / area h heght wdth. Fg. Fg. Paraol semsegment y= f( x) = h x h = area. Paraol spandrel hx y= f( x) = h = area. Fg. 5 Fg. 6 Retangle = h area. Rght trangle = h / area. Fg. 7 Fg.

6 6 V. DEMENKO MECHNCS OF MTERLS 05 Quarter rle = πr area. Quarter-rular spandrel π = r area. Fg. 9 Fg. 0 Regular hexagon sde C entrod = area. Regular hexagon hollow ross seton (syn. regular hexagon tue) t thkness = 6t area. Fg. Fg. Semrle r radus = πr area. Sne wave h = area. π Fg. Fg.

7 V. DEMENKO MECHNCS OF MTERLS 05 7 Regular polygon wth n sdes n length of a sde β entral angle for a sde α nteror angle (or vertex angle). n numer of sdes ( ) Semsegment of nth degree n n ( ) ( / ) = = ( 0) y f x h x ( / ) = h n n+ area. n > ; Fg. 5 Fg. 6 Square hmney Square ross seton square = π / area. = a area. d Fg. 7 Fg. Square tuular ross seton wdth t thkness = t area. Thn rular rng = πrt = πdt d r t << r. = ( ) Fg. 9 Fg. 0

8 V. DEMENKO MECHNCS OF MTERLS 05 Trangle h = area. Trapezod ( + ) ha = area. Fg. Fg.. Stat Moment (Frst Moment) of a Seton Consder the followng two ntegrals: S y = zd Sz = yd. () Eah of the ntegrals represents the sum of the produts of elements of area and the dstane to the respetve axs (y or z). The frst ntegral s alled the stat moment of the seton wth respet to the y axs and the seond to the z axs. The stat moment s measured n meter ued ( m ). ordng to expressons () the stat moment an e postve negatve or equal to zero. The stat moment of a ompound seton equals to the sum of the stat moments of the smplest fgures (omponents).. Central xes. Centrod Consder a plane seton and draw two pars of parallel axes y z and y z as shown n Fg.. Let a dstane etween the axes wll e and a. The area of ths seton and S y and S z are gven. t s neessary to fnd the stat moments wth respet to the y and z axes.e. S y and S z. ordng to formulas () the stat moments are S = zd y S = yd. () z

9 s may e seen from Fg. V. DEMENKO MECHNCS OF MTERLS 05 9 z = z y = y a. () Susttutng y and z from expressons () to formulas () we fnd Beause as we know y ( ) S = z d= zd d z ( ). (5) S = y a d= yd a d zd= Sy yd= Sz d= (6) then we rewrte (5) as S = S y y S = S a. (7) z z Consder the frst of the expressons derved aove: S = S. y y The quantty may e any numer whatever ether postve or negatve. t an therefore always e hosen to make the produt equal to S y. Then the stat moment wth respet to the y -axs vanshes that s 0 Sy = 0 = Sz a. () n axs wth respet to whh the stat moment s zero s alled entral axs or entrodal axs. The pont of nterseton of entral axes s alled the enter of gravty or entrod of ross-seton. Thus equatons () make t possle to determne the poston of the entrod f the stat moments are known: Sy S = Z = a= Y z =. (9)

10 0 V. DEMENKO MECHNCS OF MTERLS 05 where Z and Y are oordnates of the entrod. t s possle also to fnd the stat moments f the poston of the entrod s known. The entrod of a omposte seton s determned y Fg. Y = n n z n Z = n y (0) where y and z are oordnates of the geometral enters of the omponent fgures. Consder the smplest example. Example The alulaton of entrod oordnate of the trangular (Fg. ) Gven: s the ase of the trangle h s the heght. R.D.: dstane of the entrod of the trangle from ts ase.e. z. Fg. Sy Soluton By defnton z =. The trangle stat moment wth respet to the y axs s equal to S y = zd. n our ase d= ( z) dz = h. From trangles smlarty (z) equals to Thus ( z) = ( h z). h h h h Sy = ( h z) = 6 0 z h + 6 h = =+. h

11 V. DEMENKO MECHNCS OF MTERLS 05 The oordnate from the ase of the trangle to the entrod of gravty s h / (up dreted) and the entrodal horzontal entral axs s loated upwards at dstane / of alttude from ts ase. Example Centrodal axes of rght trangle (Fg. 5) the heght. Gven: s the ase of the trangle h s R.D.: dstane of the entrod of the trangle from ts ase.e. x and y. Fg. 5 xd Sy x 0 = = = d= hxdx ( ) ; = hx ( ) ( x) h = = hx ( ) = ( x) h h h 0 ( ) xxdx h = = = 6. h h h = Sx h By analogy y = =. n result x y axes are entrodal axes of rght trangle. Example Centrod of a omposte area (Fg. 6) seton. Gven: dmensons of an angular R.D.: x y oordnates. Fg. 6 Soluton The areas and frst moments of omposte areas may e alulated y summng the orrespondng propertes of the omponent parts. Let us assume that a omposte area s dvded nto a total of n

12 V. DEMENKO MECHNCS OF MTERLS 05 parts and let us denote the area of the th part as. Then we an otan the area and frst moments y the followng summatons: n = () = S n = y x = S n = x ; () y = n whh x and y are the oordnates of the entrod of the th part. The oordnates of the entrod of the omposte area are x Sy = = n = n = x S y x = = = n n = y. () Sne the omposte area s represented exatly y the n parts the preedng equatons gve exat results for the oordnates of the entrod. To llustrate the use of Eq. () onsder the L -shaped area (or angle seton) shown n Fg. 6 a. Ths area has sde dmensons and and thkness t. The area an e dvded nto two retangles of areas and wth entrods C and C respetvely (Fg. 6 ). The areas and entrodal oordnates of these two parts are = t ( ) = t t t x = x = y = ; t t y =. Therefore the area and frst moments of the omposte area (from Eqs. () and ()) are ( ) = + = t + t ( ) t Sx = y + y = + t t ( ) t Sy = x + x = t+ t.

13 V. DEMENKO MECHNCS OF MTERLS 05 Fnally we an otan the oordnates x and area (Fg. 6 ) from Eq. (): x Sy t+ t = = t y y of the entrod C of the omposte Sx + t t = = +. () ( + ) ( t) Note : When a omposte area s dvded nto only two parts the entrod C of the entre area les on the lne jonng the entrods C and C of the two parts (as shown n Fg. for the L-shaped area). Note : When usng the formulas for omposte areas (Eqs. () () and ()) we an handle the asene of an area y sutraton. Ths proedure s useful when there are utouts or holes n the fgure. Example Determnaton the oordnates of the entrod of the ompound seton (Fg. 7) y z h = 0mm. Gven: = 0 mm = 0 mm R.D.: x y oordnates. Soluton Dvde the area nto two smplest fgures: the rght trangle and the retangle for whh the entrods are know. Selet an artrary referene system of axes y and z and determne the oordnates of the entrod usng equaton (0). Susttutng the numeral values nto the foregong expresson we reeve: h y + h + z z z + h + h h h h z + + h + Sy Sy Sy + h + h S S + S = = = = =... Fg. 7 + = = = = =...

14 V. DEMENKO MECHNCS OF MTERLS 05 Centrods of smple fgures Crular setor Orgn of axes at enter of rle: α π α =angle n radans ( ) = αr x = rsnα rsnα y =. α Fg. Crular segment Orgn of axes at enter of rle: α π α =angle n radans ( ) = r ( α snαosα) y r sn α =. α snαosα Fg 9 soseles trangle Orgn of axes at entrod: h = x = h y =. Fg 0 Paraol semsegment paraol semsegment OB s ounded y the x axs the y axs and a paraol urve havng ts vertex at (Fg. ). The equaton of the urve s y= f ( x) = h x ()

15 V. DEMENKO MECHNCS OF MTERLS 05 5 n whh s the ase and h s the heght of the semsegment. Loate the entrod C of the semsegment. To determne the oordnates equatons: x and x y of the entrod C (Fg. ) we wll use Sy S = y x =. We egn y seletng an element of area d n the form of a thn vertal strp of wdth dx and heght y. The area of ths dfferental element s Therefore the area of the paraol semsegment s d ydx h x = = dx. () x h = d= h dx = 0 ( ). () Note: Ths area s / of the area of the surroundng retangle. Fg. Fg. The frst moment of an element of area d wth respet to an axs s otaned y multplyng the area of the element y the dstane from ts entrod to the axs. Sne the x and y oordnates of the entrod of the element shown n Fg. are x and y / respetvely the frst moments of the element wth respet to the x and y axes are

16 6 V. DEMENKO MECHNCS OF MTERLS 05 y h x h Sx = d= dx = 5 0 () x h Sy = xd= hx dx = 0 n whh we have susttuted for d from Eq. (). We an now determne the oordnates of the entrod C: (5) x y Sy = = (6) Sx h = =. (7) 5 Notes: The entrod C of the paraol semsegment may also e loated y takng the element of area d as a horzontal strp of heght dy and wdth y x=. () h Ths expresson s otaned y solvng Eq. () for x n terms of y. nother posslty s to take the dfferental element as a retangle of wdth dx and heght dy. Then the expressons for nstead of sngle ntegrals. S x and S y are n the form of doule ntegrals Fg Paraol spandrel Orgn of axes at vertex O: hx y= f ( x) = h = x = h y =. 0

17 V. DEMENKO MECHNCS OF MTERLS 05 7 Quarter rle Orgn of axes at enter of rle O: x = πr r = y =. π Fg Quarter-rular spandrel Orgn of axes at pont of tangeny: π = r r x= r π y ( ) ( 0 π ) ( π ) r = 0. r. Fg 5 Retangle Orgn of axes at entrod: = h x = h y =. Fg 6

18 V. DEMENKO MECHNCS OF MTERLS 05 Rght trangle Orgn of axes at entrod: h = x = h y =. Fg 7 Semrle Orgn of axes at entrod: = πr r y =. π Fg Fg 9 Sne wave Orgn of axes at entrod: h = π π h y =. Thn rular ar Orgn of axes at enter of rle. pproxmate formulas for ase when t s small: β angle n radans ( β π ) ; = βrt rsn β y =. β Fg 50

19 V. DEMENKO MECHNCS OF MTERLS 05 9 Centrod of a trapezod Orgn of axes at entrod: ha ( + ) = h( a+ ) y =. a+ ( ) Fg 5 Centrod of a trangle Orgn of axes at entrod: h = + x = h y =. Fg 5. xal Moments (Seond Moments) and Produt of nerta Take a ross seton of a rod. Relate t to a system of o-ordnates y z. solate an element d from the area wth o-ordnates y z. n addton to the stat moment onsder the followng four ntegrals: y = z d z = y d () yz ρ = yzd () = ρ d () Fg. 5 where the frst two ntegrals () are alled the axal moments of nerta of the seton wth respet to the y and z axes respetvely.

20 0 V. DEMENKO MECHNCS OF MTERLS 05 The thrd ntegral () s alled the produt of nerta of the seton wth respet to two mutually perpendular axes y and z. The fourth ntegral () s alled the polar moment of nerta of the seton. The dmenson of the moments of nerta s m (meters n a power of four). The axal and polar moments of nerta are always postve and annot e equal to zero. The produt of nerta may e postve negatve or equal to zero dependng on the poston of the axes. For example ths value wth respet to any par of axes s zero when ether of the axes s an axs of symmetry. Example 5 The alulaton of the axal moment of nerta of a retangle wth respet to the axes y and z passng through the entrod of the retangle (Fg. 5). Gven: and h ase and heght of the retangle respetvely. retangle. R.D.: entral axal moments of nerta of a Soluton Let us separate an elementary area d wth the ase and the heght dz at the dstane z from the axs. Sne d=dz then Fg. 5 h + h y = z d= zdz =. h The moment of nerta wth respet to the z-axs s found y a smlar way: + h z = y d= y hdy =.

21 V. DEMENKO MECHNCS OF MTERLS 05 Example 6 Calulaton of the entral moment of nerta of a rular shape (Fg. 55). Gven: d dameter of the rle. R.D.: entral axal moments of nerta. Soluton We take d as πρdρ. Thus d πd ρ = ρ d= πρ dρ =. 0 Referrng to Fg 5 we fnd ρ = y + z. ( ) That s ρ y z. Usng the symmetry we an wrte = d= y + z d= + ρ y πd = z =. 6 Example 7 Calulaton of axal moments and produt of nerta for rght trangle relatve to axs ondent wth trangle legs (Fg. 56). h z df z ( ) Fg. 55 fter susttuton Fg. 56 dz z ( ) y Gven: h R.D.: y Soluton z yz (a) Calulaton of axal moments of nerta s prelmnary determned y F = z d where d= z ( ) dz. Usng smlarty ondton z = h z ( z ) = z h h. h h z z z h h h 0 0 y = z dz = =.

22 V. DEMENKO MECHNCS OF MTERLS 05 Thus h y = y analogy h z =. () Calulaton of produt of nerta t s known that for produt of nerta yz = yzd (a) F where d= ydz. () Equaton of nlned oundary АВ s Fg. 57 z h y + = where z = h y or y z = h. () fter ths equaton (a) may e rewrtten: yz n result z z h h h h y h z h = yzdy dz = z dz = z dz = h h h 0 z z z h = + =. h yz =+. (d) Note that the propertes of strutural elements suh as hannels angles or - eams are gven n the tales of standard seton (assortments). For some geometr fgures entral moments of nerta are presented elow.

23 V. DEMENKO MECHNCS OF MTERLS 05 ssortments of steel produts Geometral propertes of angle setons wth equal legs (L shapes) (GOST 509-7) wdth of we d thkness moment of nerta radus of gyraton z 0 dstane to entrod. Desgnaton (numer) Fg. 5 d xes rea X X X0 X0 Y0 Y0 mm m x 0 max x x x 0 max y 0 mn y m сm m сm сm 0 mn сm z 0 сm Mass per meter kg

24 V. DEMENKO MECHNCS OF MTERLS 05 (fnshed)

25 V. DEMENKO MECHNCS OF MTERLS 05 5 Geometral propertes of angle setons wth unequal legs (L shapes) (GOST 50-7) B wdth of larger leg wdth of smaller leg d thkness of legs moment of nerta radus of gyraton x 0 y 0 dstanes from the entrod to the ak of the legs. Desgna ton (nume r) 5/6 / /5 5/ 5/ 56/6 6/0 7/5 75/5 /5 9/56 0/6 Fg. 59 B mm d xes rea X X Y Y 0 y 0 m x y x tan α y u mn u mn сm сm m сm m сm m сm Mass per meter kg

26 6 /7 5/ /9 6/0 V. DEMENKO MECHNCS OF MTERLS 05 (fnshed) / 0/5 5/ Geometral propertes of hannel setons (C shapes) (GOST 0-7) Fg. 60 Dmensons mm Desgnaton (numer) h s t rea m x m h heght of a eam wdth of a flange s thkness of a we t average thkness of a flange W setonal modulus radus of gyraton S x frst moment of area moment of nerta x 0 dstane from the entrod to the ak of the we. W x m x сm S x m y m W y m y сm x 0 сm Weght per meter kg

27 V. DEMENKO MECHNCS OF MTERLS 05 7 (fnshed) а а а 0 0а а а Desgnat on (numer ) 0 6 a 0 0а а Geometral propertes of S shapes (-eam setons) (GOST 9-7) Fg. 6 Dmensons mm h s t rea m x m h heght of a eam wdth of a flange s thkness of a we t average thkness of a flange axal moment of nerta W setonal modulus radus of gyraton S x frst moment of a half-seton. W x m x сm S x m y m W y m y сm Mass per meter kg

28 V. DEMENKO MECHNCS OF MTERLS 05 (fnshed) а 7 7а 0 0а Centrodal axal moments of nerta for smple fgures Crle Orgn of axes at enter of rle: πd = π r = x = 0 xy πr πd = y = = 6 p = πr = πd 5 5 x = πr = πd. 6 Fg. 6 Crle wth ore removed Orgn of axes at enter of rle: α =angle n radans ( α π /); α = aros a r = r a ; Fg. 6 x a = r α r r a a x = α 6 r r r a a = α r r xy = 0.

29 V. DEMENKO MECHNCS OF MTERLS 05 9 Fg. 6 Crular setor Orgn of axes at enter of rle: α =angle n radans ( α π /); rsnα = αr x = rsnα y = ; α r x = ( α + snαos α) r y = ( α snαos α) = = 0 xy xy αr ρ =. Fg. 65 Fg. 66 Crular segment Ellpse Orgn of axes at enter of rle: α =angle n radans ( α π /); r sn α y = α snαosα r x = ( α snαosα + sn α) = = 0 xy r y = (α snαosα sn αos α) xy Orgn of axes at entrod: = π a x π a πa = y = ; = 0 xy πa p = ( + a ). Crumferene π[.5( a+ ) a] ( a/ a).7 / a+ a (0 a/).

30 0 V. DEMENKO MECHNCS OF MTERLS 05 Fg. 67 Fg. 6 Fg. 69 Fg. 70 soseles trangle Paraol semsegment Orgn of axes at entrod: h h = x = y = ; h h x = y 6 = xy = 0 ; h ρ = (h ) + h x =. Note: For an equlateral trangle h= /. Orgn of axes at orner: y f( x) h x = = h h = x = y = ; 5 6h x = 05 Paraol spandrel Quarter rle h y = 5 Orgn of axes at vertex: h xy =. hx y= f( x) = h h = x = y = ; 0 h x = h y = 5 h xy =. Orgn of axes at enter of rle: = πr x x πr = y = 6 r = y = ; π r xy = ; (9π 6) r x = 0.05 y = r. π

31 V. DEMENKO MECHNCS OF MTERLS 05 Fg. 7 Quarter-rular spandrel Retangle Orgn of axes at pont of tangeny: π = r r x = r ( π ) (0 π ) r y = 0.r ( π ) 5π x = r 0.05r 6 π y = x = r 0.70r 6. a Fg 7 a) Orgn of axes at entrod: h = h x = y = ; h h x = y = xy = 0 ; h ρ = ( h ) +. ) Orgn of axes at orner: h x = h h y = xy = ; h ρ = ( h ) + x h =. 6( + h )

32 V. DEMENKO MECHNCS OF MTERLS 05 Fg. 7 Regular polygon wth n sdes Orgn of axes at entrod: C = entrod (at enter of polygon) n = numer of sdes ( n ) = length of a sde β = entral angle for a sde α = nteror angle (or vertex angle) 60 n β = α = 0 n n α + β = 0 ; R = radus of rumsred rle (lne C) R = radus of nsred rle (lne CB) β β R = s R = ot n β = ot ; moment of nerta aout any axs through C (the entrod C s a prnpal pont and every axs through C s a prnpal axs) Rght trangle n β β = ot ot + 9 ρ =. a Fg. 7

33 V. DEMENKO MECHNCS OF MTERLS 05 a) Orgn of axes at entrod: h h = x = y = ; h h x = y 6 = 6 = ( h)/7 ; xy h ρ = ( + ) h x h 6 Fg. 75 =. Semrle ) Orgn of axes at vertex: h x = h y = h p = ( h + ) h xy = ; h x =. Orgn of axes at entrod: = π r y r = ; π (9π 6) r = 0.09r 7π y = π r = 0 xy = x = πr. x xy Fg. 76 Fg. 77 Semsne wave Thn rular ar Orgn of axes at entrod: = h y π = π h ; x = ( π ) h h 9π 6 y = h 0.h π π xy = 0 xy = h x =. 9 π Orgn of axes at enter of rle. pproxmate formulas for ase when t s small: β angle n radans ( β π /) ; = βrt y = ( rsn β)/ β ; x = rtβ ( + sn β os β ) y = rtβ ( sn β os β ) = = 0 xy β + snβ osβ x = rt β. Note: For a semrular ar ( β = π /). xy

34 V. DEMENKO MECHNCS OF MTERLS 05 Thn rular rng Orgn of axes at entrod. pproxmate formulas for ase when t s small: = πrt = πdt x y πrt πdt = = = = 0 ρ xy = πrt= πdt. Fg. 7 Thn retangle Orgn of axes at entrod. pproxmate formulas for ase when t s small: = t x y x = t sn β = t os β = t sn β. Fg. 79 Fg. 0 Trapezod Orgn of axes at entrod: ha ( + ) = h( a+ ) y = ( a+ ) x h ( a + a+ ) = 6( a+ ) h ( a+ ) x =.

35 V. DEMENKO MECHNCS OF MTERLS 05 5 Trangle a Fg. a) Orgn of axes at entrod: h + h = x = y = ; h x = 6 h y = ( + ); 6 h xy = ( ) 7 h ρ = ( h ) ) Orgn of axes at vertex: h x = h y = ( + ) h xy = ( ) h x =.